Chapter 9, Problem 9.11P

(a)

Interpretation Introduction

Interpretation:

Equilibrium chemical reaction of imidazole and imidazole hydrochloride with constants $K_{\text{b}}$ and $K_{\text{a}}$ has to be determined.

Concept Introduction:

$K_{\text{a}}$ for reaction is defined as dissociation constant for acid. It helps to determine the amount of hydrogen ions in acid. $K_{\text{b}}$ for reaction is defined as association constant for base. It helps to determine the amount of hydroxide ions in base.

Explanation of Solution

The equilibrium reaction for imidazole with equilibrium constant $K_{\text{b}}$ is as follows:

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}{\text{N}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}\stackrel{K_{\text{b}}}{\rightleftharpoons }{\text{C}}_{\text{3}}{\text{NH}}_{\text{4}}{\text{NH}}^{+}+{\text{OH}}^{-}$

The equilibrium reaction for imidazole hydrochloride with equilibrium constant $K_{\text{a}}$ is as follows:

  ${\text{C}}_{\text{3}}{\text{NH}}_{\text{4}}{\text{NH}}^{+}{\text{Cl}}^{-}+{\text{H}}_{\text{2}}\text{O}\stackrel{K_{\text{a}}}{\rightleftharpoons }{\text{C}}_{\text{3}}{\text{N}}_2{\text{H}}_{\text{4}}{}^{+}+{\text{H}}_3{\text{O}}^{+}{\text{Cl}}^{-}$

(b)

Interpretation Introduction

Interpretation:

Value of $\text{pH}$ for a solution that contains imidazole and imidazole hydrochloride has to be calculated.

Concept Introduction:

$K_{\text{a}}$ for reaction is defined as dissociation constant for acid. It helps to determine the amount of hydrogen ions in acid. $K_{\text{b}}$ for reaction is defined as association constant for base. It helps to determine the amount of hydroxide ions in base.

The expression used to calculate concentration of compound is as follows:

  $\text{Concentration}=\frac{\text{moles}}{\text{volume}\left(\text{L}\right)}$

The formula for moles is as follows:

  $\text{Moles}=\frac{\text{mass of solute}}{\text{molar mass of solute}}$

Explanation of Solution

The formula to calculate concentration of imidazole $\left(\text{I}\right)$ is as follows:

  $\left[\text{I}\right]=\frac{\left(\frac{m_{\text{I}}}{M_{\text{I}}}\right)}{V\left(\text{L}\right)}$        (1)

Here,

$\left[\text{I}\right]$ is concentration of imidazole.

$m_{\text{I}}$ is mass of imidazole.

$M_{\text{I}}$ is molar mass of imidazole.

$V$ is volume of solution.

Substitute $1\text{ g}$ for $m_{\text{I}}$, $\text{68}\text{.08 g/mol}$ for $M_{\text{I}}$ and $0.1\text{ L}$ for $V$ in equation (1).

  $\begin{array}{c}\left[\text{I}\right]=\frac{\left(\frac{1\text{ g}}{\text{68}\text{.08 g/mol}}\right)}{0.1\text{ L}}\\ =0.1469\text{ mol/L}\end{array}$

The formula to calculate concentration of imidazole hydrochloride $\left({\text{IH}}^{+}\right)$ is as follows:

  $\left[{\text{IH}}^{+}\right]=\frac{\left(\frac{m_{{\text{IH}}^{+}}}{M_{{\text{IH}}^{+}}}\right)}{V\left(\text{L}\right)}$        (2)

Here,

$\left[{\text{IH}}^{+}\right]$ is concentration of imidazole hydrochloride.

$m_{{\text{IH}}^{+}}$ is mass of imidazole hydrochloride.

$M_{{\text{IH}}^{+}}$ is molar mass of imidazole hydrochloride.

$V$ is volume of solution.

Substitute $1\text{ g}$ for $m_{{\text{IH}}^{+}}$, $\text{104}\text{.54 g/mol}$ for $M_{{\text{IH}}^{+}}$ and $0.1\text{ L}$ for $V$ in equation (2).

  $\begin{array}{c}\left[{\text{IH}}^{+}\right]=\frac{\left(\frac{1\text{ g}}{\text{104}\text{.54 g/mol}}\right)}{0.1\text{ L}}\\ =0.0966\text{ mol/L}\end{array}$

The expression to calculate $\text{pH}$ is as follows:

  $\text{pH}=\text{p}{K}_a+\log \frac{\left[\text{I}\right]}{\left[{\text{IH}}^{+}\right]}$        (3)

Here,

$\text{p}{K}_a$ is dissociation constant of acid.

$\left[\text{I}\right]$ is concentration of imidazole.

$\left[{\text{IH}}^{+}\right]$ is concentration of imidazole hydrochloride.

Substitute 7 for $\text{p}{K}_a$, $0.1469\text{ mol/L}$ for $\left[\text{I}\right]$ and $0.0966\text{ mol/L}$ for $\left[{\text{IH}}^{+}\right]$ in equation (3).

  $\begin{array}{c}\text{pH}=7+\log \frac{\left(0.1469\text{ mol/L}\right)}{\left(0.0966\text{ mol/L}\right)}\\ =7+0.18\\ =7.18\end{array}$

Hence, value of $\text{pH}$ is 7.18.

(c)

Interpretation Introduction

Interpretation:

Value of $\text{pH}$ if ${\text{HClO}}_4$ is added to solution has to be calculated.

Concept Introduction:

$K_{\text{a}}$ for reaction is defined as dissociation constant for acid. It helps to determine the amount of hydrogen ions in acid. $K_{\text{b}}$ for reaction is defined as association constant for base. It helps to determine the amount of hydroxide ions in base.

The expression used to calculate concentration of compound is as follows:

  $\text{Concentration}=\frac{\text{moles}}{\text{volume}\left(\text{L}\right)}$

The formula for moles is as follows:

  $\text{Moles}=\frac{\text{mass of solute}}{\text{molar mass of solute}}$

Explanation of Solution

The formula to calculate concentration of ${\text{HClO}}_4$ is as follows:

  $\left[{\text{HClO}}_4\right]=\frac{n_{{\text{HClO}}_4}}{V\left(\text{L}\right)}$        (4)

Rearrange equation (4) for $n_{{\text{HClO}}_4}$.

  $n_{{\text{HClO}}_4}=\left[{\text{HClO}}_4\right]\left(V\left(\text{L}\right)\right)$        (5)

Here,

$\left[{\text{HClO}}_4\right]$ is concentration of ${\text{HClO}}_4$.

$n_{{\text{HClO}}_4}$ ismoles of ${\text{HClO}}_4$.

$V$ is volume of ${\text{HClO}}_4$.

Substitute $1.07\text{ M}$ for $\left[{\text{HClO}}_4\right]$ and $2.30\text{ mL}$ for $V$ in equation (5).

  $\begin{array}{c}{n}_{{\text{HClO}}_4}=\left(1.07\text{ M}\right)\left(2.30\text{ mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =0.0025\text{ mol}\end{array}$

On addition of ${\text{HClO}}_4$ moles of imidazole reduce by $0.0025\text{ mol}$. Hence, number of moles of imidazole is as follows:

  $\begin{array}{c}{n}_{\text{imidazole}}=0.1496\text{ mol}-0.0025\text{ mol}\\ =0.0147\text{ mol}\end{array}$

On addition of ${\text{HClO}}_4$ moles of imidazole hydrochloride is increased by $0.0025\text{ mol}$. Hence, number of moles of imidazole hydrochloride is as follows:

  $\begin{array}{c}{n}_{\text{imidazole hydrochloride}}=0.0966\text{ mol}+0.0025\text{ mol}\\ =0.098\text{ mol}\end{array}$

Since volume is same, therefore, number of moles of imidazole and imidazole hydrochloride is equal to its moles. Hence, value of $\left[\text{I}\right]$ is $0.147\text{ M}$ and $\left[{\text{IH}}^{+}\right]$ is $0.098\text{ M}$.

The expression to calculate $\text{pH}$ is as follows:

  $\text{pH}=\text{p}{K}_a+\log \frac{\left[\text{I}\right]}{\left[{\text{IH}}^{+}\right]}$        (6)

Here,

$\text{p}{K}_a$ is dissociation constant of acid.

$\left[\text{I}\right]$ is concentration of imidazole.

$\left[{\text{IH}}^{+}\right]$ is concentration of imidazole hydrochloride.

Substitute 7 for $\text{p}{K}_a$, $0.147\text{ M}$ for $\left[\text{I}\right]$ and $0.098\text{ M}$ for $\left[{\text{IH}}^{+}\right]$ in equation (3).

  $\begin{array}{c}\text{pH}=7+\log \frac{\left(0.147\text{ M}\right)}{\left(0.098\text{ M}\right)}\\ =7+0.1\\ =7.1\end{array}$

Hence, value of $\text{pH}$ is 7.1.

(d)

Interpretation Introduction

Interpretation:

Volume in milliliters of ${\text{HClO}}_4$ added to imidazole $\left(\text{I}\right)$ has to be calculated.

Concept Introduction:

The equation for buffer is described by Henderson-Hasselbach equation and it is a rearranged for of equilibrium constant, $K_{\text{a}}$. Consider an equation of dissociation of an acid as follows:

  $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

Formula to calculate $\text{pH}$ is as follows:

  $\text{pH}=p{K}_{\text{a}}+\log \left(\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\right)$

Here,

$p{K}_{\text{a}}$ is acid dissociation constant of weak acid $\text{HA}$.

$\left[{\text{A}}^{-}\right]$ is concentration of conjugate base ${\text{A}}^{-}$

$\left[\text{HA}\right]$ is concentration of acid $\text{HA}$.

The expression used to calculate concentration of compound is as follows:

  $\text{Concentration}=\frac{\text{moles}}{\text{volume}\left(\text{L}\right)}$

The formula for moles is as follows:

  $\text{Moles}=\frac{\text{mass of solute}}{\text{molar mass of solute}}$

Explanation of Solution

The formula to calculate number of moles of imidazole is as follows:

  $n_{\text{I}}=\frac{m_{\text{I}}}{M_{\text{I}}}$        (7)

Here,

$n_{\text{I}}$ is number of moles of imidazole.

$m_{\text{I}}$ is mass of imidazole.

$M_{\text{I}}$ is molar mass of imidazole.

Substitute $1\text{ g}$ for $m_{\text{I}}$ and $\text{68}\text{.08 g/mol}$ for $M_{\text{I}}$ in equation (7).

  $\begin{array}{c}{n}_{\text{I}}=\frac{1\text{ g}}{\text{68}\text{.08 g/mol}}\\ =0.015\text{ mol}\end{array}$

The formula to calculate concentration of ${\text{HClO}}_4$ is as follows:

  $\left[{\text{HClO}}_4\right]=\frac{n_{{\text{HClO}}_4}}{V\left(\text{L}\right)}$        (8)

Rearrange equation (8) for $n_{{\text{HClO}}_4}$.

  $n_{{\text{HClO}}_4}=\left[{\text{HClO}}_4\right]\left(V\left(\text{L}\right)\right)$        (9)

Here,

$\left[{\text{HClO}}_4\right]$ is concentration of ${\text{HClO}}_4$.

$n_{{\text{HClO}}_4}$ is moles of ${\text{HClO}}_4$.

$V$ is volume of ${\text{HClO}}_4$.

Substitute $1.07\text{ mol/L}$ for $\left[{\text{HClO}}_4\right]$ and $x\text{ L}$ for $V$ in equation (9).

  $\begin{array}{c}{n}_{{\text{HClO}}_4}=\left(1.07\text{ mol/L}\right)\left(x\text{ L}\right)\\ =1.07x\text{ mol}\end{array}$

The ICE table to calculate concentration is as follows:

  $\begin{array}{cccccc}& \text{I}& +& {\text{HClO}}_4& \to & {\text{IH}}^{+}\\ \text{Initial moles}& 0.0147& & 1.07x& & -\\ \text{Change in moles}& -1.07x& & -& & 1.07x\\ \text{Final moles}& 0.0147-1.07x& & -& & 1.07x\end{array}$

In accordance to Henderson Hasselbach equation, $\text{pH}$ is calculated as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{\left[\text{I}\right]}{\left[{\text{IH}}^{+}\right]}$        (10)

Substitute 7 for $\text{p}{K}_a$, $0.0147-1.07x$ for $\left[\text{I}\right]$ and $1.07x$ for $\left[{\text{IH}}^{+}\right]$ in equation (10).

  $\begin{array}{c}\text{pH}=7+\log \frac{\left(0.0147-1.07x\right)}{\left(1.07x\right)}\\ 6.993=7+\log \frac{\left(0.0147-1.07x\right)}{\left(1.07x\right)}\\ x=0.007\end{array}$

Hence, volume of solution is as follows:

  $\begin{array}{c}\text{Volume}=0.007\text{ L}\left(\frac{1000\text{ mL}}{1\text{ L}}\right)\\ =6.86\text{ mL}\end{array}$.

Hence, volume is $6.86\text{ mL}$.